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Is there any easy way to see the following?

Let $n\in\mathbb{N}$. Let $a,b\in\mathbb{R}$ s.t. $a < b$. Then, there exists $q\in\mathbb{Q}$ s.t. $a<q<b$ and $p(x) = x^n - q$ is irreducible over $\mathbb{Q}[x]$.

That is, given a degree $n$ and an open real interval $I$, we can always find some rational $q\in I$ s.t. the polynomial $p(x) = x^n - q$ is irreducible over $\mathbb{Q}[x]$.

My current solution (using the binomial irreducibility criteria given here - Is the polynomial $x^{105} - 9$ reducible over $\mathbb{Z}$?) seems like overkill.

Thank you!

Community
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letsmakeitwork42
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2 Answers2

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Probably an overkill:

I will prove the following statement: If $0<a<b$ we can find some $r \in \mathbb Q$ with $a < r <b$ so that all polynomials $X^n \pm r$ are irreducible.

Using the prime number theorem you can prove that for each $\epsilon >1$ there exists some $N$ so that for all $n \geq N$ there exists a prime between $n $ and $\epsilon n$.

Now, let $\epsilon =\frac{b}{a}$ and pick some prime $q$ so that $bq > N_\epsilon$. Then, there exists some prime $p$ so that $$bq < p <\frac{a}{b}bq =aq$$

This shows that we can find some $a <r <b$ rational which is of teh form $r=\frac{p}{q}$ with $p,q$ primes.

Now, by Eisenstein criteria, the polynomials $$qX^n \pm p$$ are irreducible over $\mathbb Q$.

N. S.
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First note that we can find a rational $\frac{a}{2}<\frac{k}{l}<\frac{b}{2}$ such that $k,l$ are odd (I leave it as an exercise, you can even constrain $l$ to be e.g. large power of $3$). Then I claim that $\frac{2k}{l}$ makes this work. Indeed, if $x^n-\frac{2k}{l}$ was reducible over $\Bbb Q$, then so would be $lx^n-2k$. But by Eisenstein's criterion it isn't the case, as $2\mid 2k,2^2\nmid 2k$ (because $k$ is odd) and $2\nmid l$ ($l$ is odd).

Wojowu
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